You have found the following ages (in years) of 4 zebras. The zebras are randomly selected from the 48 zebras at your local zoo: $ 8,\enspace 13,\enspace 5,\enspace 10$ Based on your sample, what is the average age of the zebras? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 48 zebras, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{1} + {16} + {16} + {1}} {{4 - 1}} $ $ {s^2} = \dfrac{{34}}{{3}} = {11.33\text{ years}^2} $ We can estimate that the average zebra at the zoo is 9 years old. There is a variance of 11.33 years $^2$.